Evaluate $(1-\frac1{2^2})(1-\frac1{3^2})\ldots(1-\frac1{2015^2})$

Question

Evaluate $$\prod_{k=2}^{2015} \left(1-\frac1{k^2}\right) = \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\ldots\left(1-\frac{1}{2014^2}\right)\left(1-\frac{1}{2015^2}\right)$$

Answer 1

Simple operations lead to: $$\frac{(2^2-1)(3^2-1)...(2015^2-1)}{2^23^2...2015^2}$$. The numerator can be calculated using $$(a^2-1) = (a-1)(a+1)$$ and the denominator is $(2015)!^2$. The final answer is $\frac{1008}{2015}$.

Answer 2

The answer is $1008/2015$. Indeed,

$$\prod_{k = 2}^{2015} \left(1 - \frac{1}{k^2}\right) = \prod_{k = 2}^{2015} \frac{k^2 - 1}{k^2} = \prod_{k = 2}^{2015} \dfrac{\frac{k-1}{k}}{\frac{k}{k+1}} = \dfrac{\frac{2-1}{2}}{\frac{2015}{2015 + 1}} = \frac{1}{2}\frac{2016}{2015} = \frac{1008}{2015}.$$

Answer 3

I believe you should keep in mind the general thing
$$\frac{(n-1)!(n+1)!}{2(n!)^2}=\dots=\frac{n+1}{2n}$$ $$\frac{n+1}{2n}=\frac{2015+1}{2\times2015}=\frac{1008}{2015}$$

Answer 4

I saw really no effort put into the question.

Even if you are not really good at maths, you can evaluate the first few products. Always helps, just calculating a few results so you get an idea how the problem works.

The first product is 3/4.

Multiplied by 8/9 = 2/3.

Multiplied by 15/16 = 5/8.

Multiplied by 24/25 = 3/5.

Multiplied by 35/36 = 7/12.

Multiplied by 48/49 = 4/7.

Now do you see a pattern? If you multiply numerator and denominator each by 2 in every second line, you get the pattern

3/4, 4/6, 5/8, 6/10, 7/12, 8/14

Now you can write down a hypothesis, and prove it using induction. Really easy. Even for a beginner.

What you added in a comment to your question is really nonsense. Look at the product. Each of the numbers multiplied is very, very close to 1. All the numbers that you multiply in your comment are tiny. It should have been obvious that this cannot be right.